SAT, Geometry – triangles

Q: The lengths of two sides of a triangle are 4 and 6. If the third side is an integer, what is one possible value for the perimeter?

Answer:  The perimeter can be any of the following values: 13, 14, 15, 16, 17, 18, or 19.

Explanation: The length of the third side of any triangle must be longer than the difference of the other two sides, but shorter than the sum. (This is called the Triangle Inequality Theorem.) Here, the third side is longer than 6-4=2 and shorter than 6+4=10. It can be any integer between, but not including, 2 and 10. Recall that integers include only whole numbers and their negatives. The third side must then be either 3, 4, 5, 6, 7, 8, or 9. To get the possible perimeters, add the other two side lengths (4 and 6) to each of the possibilities for the third side. The result is that the perimeter can be 13, 14, 15, 16, 17, 18, or 19.

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SAT – Average speed

Q:  Martin drives 120 miles from Oaktown to Mapleville. He drives for 60 minutes, stops for lunch for 30 minutes, then completes his journey in 60 more minutes. Including his stopping time, what is Martin’s average speed in miles per hour for this trip?

Answer:  48

Explanation:  Average speed = (total distance) / (total time).

The total distance is 120 miles.

The total time in minutes is 60 + 30 + 60 = 150 minutes.

Convert this to hours:  150/60 = 2.5 hours

Average speed = (120 miles) / (2.5 hours) = 48 miles per hour

Hint:  If you didn’t know the ‘formula’, just look at the units that the problem mentions:  miles per hour.  That’s miles divided by hours.  Do the total of each amount, and divide them.

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Algebra 2 – proportions

Q:  If x is inversely proportional to the square of y, and x = 4 when y = 1, find x when y = 2.

Answer:  1

Explanation:  We are told that x and y² are inversely proportional, so x • y² = k, where k is a constant.

Let’s find out what k is, using the given info:

4 • 1² = k

k = 4

Now,  x • y² = 4.

Plug in the new value for y, to find the new value for x:

x • y² = 4

x • 2² = 4

x • 4 = 4

x = 1

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SAT – Counting and Ordering

Q:  20 students are to choose a president and vice president from among themselves. How many possible ways can they be chosen?

Answer:  380

Explanation 1:  Using the Counting Principle, set up two blanks to represent the president and vice president, from left to right:

____    ____

There are 20 choices for the president.  Put 20 in the first blank:

_20_    ____

The same person cannot be chosen for both positions, so there are 19 left to choose from for the vice president.  Put 19 in the second blank:

_20_    _19_

Now, multiply these numbers.  20 x 19 = 380.

Explanation 2:  This is a permutation problem since you are choosing 2 people from a total of 20 people, and the order is important.  The order is important because the two people chosen must be given specific positions, and switching those positions results in a different scenario.  Do  20 nPr 2 , or use the formula for nPr.  On the TI-84 Plus, find nPr in MATH-PRB-2.  The result is 380.

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SAT, Algebra 1 – percent change

Q: Last year, Emmy was 40 inches tall.  This year, she is 42 inches tall.  By what percent has her height increased?

 Answer: 5%

 Explanation:  Percent change is calculated by the following formula:

 (New – Original)/Original  x 100%

 (42 – 40)/40 = 2/40 = .05

 .05 x 100% = 5%

 Since it is positive, this is a percent increase.

My TI-84 Plus program PRCNTCHG will calculate this, too.

 

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SAT – number properties

Q: What is the sum of all of the integers from 1 to 100?

Answer: 5050

Explanation: Don’t add 1 + 2 + 3 + 4 + … all the way up to 100! While it will work, it is a giant waste of time. Look for a pattern of numbers to pair up:

 1 + 99 = 100

2 + 98 = 100

3 + 97 = 100

49 + 51 = 100

 That’s 49 one hundreds. 49×100 = 4900. Now, add the numbers we have not counted yet: 100 and 50.

4900 + 100 + 50 = 5050.

 My TI-84 Plus program ‘SUMATOB’ will find this sum, too.

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Algebra 2 – Sequences

Q: What is the tenth term of the sequence: 4, 12, 36, 108, …?

Answer:  78732

Explanation:  It’s a geometric sequence because each term is multiplied by 3 to get the next term. The common ratio (r) is 3. You *could* just keep multiplying by 3 until you get to the tenth term. It’s not the most efficient way, but it works.

The ‘best’ way, if you know it, is to use the formula for the nth term of a geometric sequence:

2014-09-06_202336

Oh, and my TI-84 Plus program ‘NTHTERM’ will find this answer for you.

 

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Geometry – Lines and segments

Q: Points D, E, and F are collinear with DE < EF and DF < EF.  If DE = 2x+6, DF = 3x-4, and EF = 22, what is DE?

 Answer: 14

 Explanation:

 First, recall that ‘DE’ refers to the length of the segment with endpoints D and E.  It is the distance between D and E.  Next, DON’T ASSUME that E is between the D and F!  Since DE and DF are each less than EF, and all three points lie on the same line (collinear), D must be between E and F.

 Draw the diagram:

__________________________

E                         D                   F

 Now label the distances as they are given in the problem:

  <————–22——————->

__________________________

E        2x+6         D     3x-4      F

 Use the concept: Part + Part = Whole (also called the Segment Addition Postulate) to write the equation:

 2x+6 + 3x-4 = 22

 Solve:

5x + 2 = 22

5x = 20

x = 4

 Plug x=4 back in to the ‘DE’ part:

DE = 2x+6 = 2(4) + 6 = 14.

 Tip:  To be sure your answer is correct, plug x=4 into the ‘DF’ part, too:

DF = 3x-4 = 3(4)-4 = 8.  Now, see that the sum of the two ‘parts’ does equal the ‘whole’:

14 + 8 = 22.

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Algebra 1, 2 – solving quadratic equations

Q: The square of a positive number is equal to 5 more than the product of 4 and the same number. What is the number?

Answer: 5

Explanation:

2014-09-04_231259

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SAT – number properties

Q: How many positive integers less than 500 are multiples of 3, 7, and 11?

Answer:  2

Explanation:  3, 7, and 11 do not have any common factors except 1.  So, their least common multiple is 3 x 7 x 11 = 231.  The next two common multiples are 231 x 2 = 462 and 231 x 3 = 693.  693 is greater than 500.  There are only two that are less than 500.  They are 231 and 462. 

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